Ore condition

The (left or right) Ore conditions are a criterion in ring theory , a subfield of algebra , which allows the formation of quotient fields or general localizations to be generalized to the case in which the underlying ring is not commutative is. They are named after their discoverer, Øystein Ore . Rings that fulfill them are called (left or right) Ore rings .

motivation

In commutative algebra , the location of rings is a useful tool. Roughly speaking, this consists of making elements of a subset of the ring invertible or "admitting them as denominators". So that this can be meaningfully defined, it is only necessary there that the quantity is multiplicative and contains 1 (usually this is also required). ${\displaystyle S}$ ${\displaystyle S}$ ${\displaystyle 0\notin S}$

As soon as one tries to generalize this approach to non-commutative rings, one encounters several problems. In the abstract one can always form a ring in which the elements of become invertible and which satisfies a suitable universal property analogous to that in the commutative case, but this generally has poor properties and is not easy to specify in concrete terms. Difficulties arise even for rings with zero divisors . For example, it has been shown that there are zero-divisor-free rings that cannot be embedded in an oblique body . In particular, there cannot be a kind of “quotient inclined body” in full generality analogous to the quotient field for areas of integrity . ${\displaystyle S}$

The Norwegian mathematician Øystein Ore gave in an article in 1931 a criterion that allows the formation of certain rings of quotients. Ore's considerations were later generalized by Keizo Asano and others.

Special case: rings free of zero divisors

Let be a ring (with 1) without a zero divisor. fulfills the right-ore condition if the following applies to all : ${\displaystyle R}$ ${\displaystyle R}$ ${\displaystyle a,b\in R-\lbrace 0\rbrace }$

${\displaystyle aR\cap bR\neq \lbrace 0\rbrace }$.

That means, and have other common multiple "from the right" besides the 0. It is then also called a right-ore-ring . ${\displaystyle a}$ ${\displaystyle b}$ ${\displaystyle R}$

Similarly, the links-ore condition is defined by for all . ${\displaystyle Ra\cap Rb\neq \lbrace 0\rbrace }$ ${\displaystyle a,b\in R-\lbrace 0\rbrace }$

Formation of "quotient oblique bodies"

If the right-ore condition is met, a quotient inclined body can be formed, similar to the formation of the quotient field . The elements are again written as fractions, such as ${\displaystyle R}$

${\displaystyle a/s}$with . ${\displaystyle a\in R,s\in R-\lbrace 0\rbrace }$

Two “breaks” and are identified if there are further elements such that and applies. (Formally, this defines an equivalence relation on the set and denotes the equivalence class of .) ${\displaystyle a/s}$ ${\displaystyle a'/s'}$ ${\displaystyle b,b'}$ ${\displaystyle sb=s'b'\in S}$ ${\displaystyle ab=a'b'}$ ${\displaystyle R\times (R-\lbrace 0\rbrace )}$ ${\displaystyle a/s}$ ${\displaystyle (a,s)}$

For these “fractions”, the addition and multiplication are now defined according to certain formulas, which are a little more complicated than the usual rules for calculating fractions. For the definitions (as well as for the fact that the above identification was actually an equivalence relation) the right-ore condition must be used.

The addition and multiplication defined in this way actually turn the set of those “fractions” into an inclined body , and the mapping defines an embedding of after . ${\displaystyle Q^{r}(R)}$ ${\displaystyle a\mapsto a/1}$ ${\displaystyle R}$ ${\displaystyle Q^{r}(R)}$

In addition, the following universal property applies: If a ring homomorphism is such that there is a unit in for all , then it clearly continues to a ring homomorphism . ${\displaystyle \alpha \colon R\rightarrow R'}$ ${\displaystyle \alpha (a)}$ ${\displaystyle a\neq 0}$ ${\displaystyle R'}$ ${\displaystyle \alpha }$ ${\displaystyle {\hat {\alpha }}\colon Q^{r}(R)\rightarrow R'}$

Similarly, everything can be defined “from the left”. It should be noted that a ring can meet the left-ore condition without being a right-ore-ring, and vice versa (see examples). However, if a ring is both a left and a right Ore-ring (one simply says "Ore-Ring"), the associated left and right quotient oblique bodies are isomorphic.

Properties and examples

• Every (left / right) Noetherian zero divider ring fulfills the (left / right) Ore condition.
• A ring free of zero divisors is a (left / right) Ore ring if and only if it is uniform over itself as a (left / right) module, ie two non-trivial sub-modules each have a non-trivial section.
• The ring of integer quaternions is an Ore-ring and has that of the rational quaternions as a quotient oblique.
• Be and the Frobenius homomorphism (dh ). Then the ring of skew polynomials is a zero-divisional left-ore-ring, but not a right-ore-ring. ${\displaystyle k=\mathbb {F} _{p}(t)}$ ${\displaystyle \sigma \colon k\rightarrow k}$ ${\displaystyle \sigma (a)=a^{p}}$ ${\displaystyle R=k[x;\sigma ,0]}$

Ore rings

Now be any non-commutative ring. Left or right zeros can occur and these cannot be reasonably accepted as denominators. Instead, the set of all regular elements (ie those that are neither left nor right zeros) can be used as the denominator . This is multiplicative, contains the 1 but not the 0. In the special case above, was . ${\displaystyle R}$ ${\displaystyle S}$ ${\displaystyle S=R-\lbrace 0\rbrace }$

${\displaystyle R}$satisfies the right-ore condition if for all elements exist such that ${\displaystyle s\in S,a\in R}$ ${\displaystyle t\in S,r\in R}$

${\displaystyle at=sr}$

or equivalent:

${\displaystyle aS\cap sR\neq \emptyset }$.

(One can easily show that in the above special case this is equivalent to the condition given there.)

A ring that meets the right-ore condition is called a right-ore ring . By turning all products around, the analogous definitions for the left-ore condition and left-ore rings are obtained .

Ring of (right) quotients

We now want to construct a ring of right quotients as well as an injective ring homomorphism that should meet the following conditions: ${\displaystyle RS^{-1}}$ ${\displaystyle \phi \colon R\rightarrow RS^{-1}}$

• There is a unity for everyone . ${\displaystyle s\in S}$ ${\displaystyle \phi (s)}$
• Each element of can be written as having appropriate . ${\displaystyle RS^{-1}}$ ${\displaystyle \phi (a)\phi (s)^{-1}}$ ${\displaystyle a\in R,s\in S}$

Again, analog definitions “from the left” are possible, then you write . ${\displaystyle S^{-1}R}$

The set of Ore is an exact criterion for when there is such a ring of quotients:

${\displaystyle R}$has an embedding in a ring of right quotients if and only if is a right ore ring. ${\displaystyle \phi \colon R\rightarrow RS^{-1}}$ ${\displaystyle R}$

${\displaystyle RS^{-1}}$is also called here the "classic ring of right quotients" and is denoted by. (Similarly, everything "from the left" with the label .) ${\displaystyle Q_{cl}^{r}(R)}$ ${\displaystyle Q_{cl}^{l}(R)}$

If both a left- and a right-Ore-ring, the corresponding classical rings of left or right quotients are isomorphic : . ${\displaystyle R}$ ${\displaystyle Q_{cl}^{l}(R)\cong Q_{cl}^{r}(R)}$

Properties and examples

• Every commutative ring is an ore ring. (All left / right properties coincide, and the usual localization is the ring of quotients.)
• Let be a field, the polynomial ring in the variable and the ring of the rational functions over in . Then the ring ${\displaystyle k}$ ${\displaystyle k[x]}$ ${\displaystyle x}$ ${\displaystyle k(x)}$ ${\displaystyle k}$ ${\displaystyle x}$

${\displaystyle R={\begin{pmatrix}k&k[x]\\0&k[x]\end{pmatrix}}}$ a right ore ring with a classic ring of right quotients , but R is not a left ore ring. For example , ie the links-ore-condition is violated. ${\displaystyle R={\begin{pmatrix}k&k(x)\\0&k(x)\end{pmatrix}}}$ ${\displaystyle S\cdot {\begin{pmatrix}0&1\\0&0\end{pmatrix}}\cap R\cdot {\begin{pmatrix}1&0\\0&x\end{pmatrix}}=\emptyset }$

Further generalization

The above definition of a ring of (right) quotients can be easily modified and transferred to more general ones (in contrast to the “classic” regular elements of ). In general, however, we can then no longer require that it be injective. A reasonable substitute for this is the additional condition: ${\displaystyle RS^{-1}}$ ${\displaystyle S}$ ${\displaystyle S=}$ ${\displaystyle R}$ ${\displaystyle \phi }$

• ker . ${\displaystyle \phi =\lbrace a\in R:as=0\;{\mbox{ für geeignetes}}\;s\in S\rbrace }$

It turns out that such a ring of (right) quotients with respect to can be formed if and only if it fulfills the following properties: ${\displaystyle S}$ ${\displaystyle S}$

• Elements exist for all such that . (This is just the generalization of the right-ore condition for the set .) ${\displaystyle s\in S,a\in R}$ ${\displaystyle t\in S,r\in R}$ ${\displaystyle at=sr}$ ${\displaystyle S}$
• Be . If there is a with , then there is also a with . (This condition was previously empty because it only consisted of regular elements.) ${\displaystyle a\in R}$ ${\displaystyle s\in S}$ ${\displaystyle sa=0}$ ${\displaystyle s'\in S}$ ${\displaystyle as'=0}$ ${\displaystyle S}$

swell

1. ↑ Lam, p. 292, Theorem 9.11. (The example is from AI Malzew from 1937.)
2. ↑ Øystein Ore : Linear equations in non-commutative fields. In: Annals of Mathematics. 32, 1931, ISSN  0003-486X , pp. 463-477.
3. ↑ Keizo Asano: About the formation of quotients in oblique rings. In: Journal of the Mathematical Society of Japan. Vol. 1, No. 2, 1949, ISSN  0025-5645 , pp. 73-78, doi : 10.2969 / jmsj / 00120073 .

literature

• TY Lam : Lectures on Modules and Rings. Springer-Verlag, Berlin et al. 1999, ISBN 0-387-98428-3 ( Graduate Texts in Mathematics 189).